Dec 2015 Paper II - Part-2

Introduction

The UGC NET JRF Computer Science is a great way to start your career as a Computer Science Lecturer. The UGC's NET exams provide an opportunity to begin a career as a lecturer in colleges, universities, and other educational institutions. The tests are held twice a year. 

Note: This article contains Q.No. 26 to Q.No. 50 out of the 50 questions asked in UGC NET Dec 2015 paper II. The solutions to Q.No. 1 to Q.No. 25 can be found in the UGC NET Dec 2015 Part-1 article.
 

Question 1: A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries:

Page	Block
0	3
2	1
5	2
7	0

Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU?

1. 1024, 3072, 4096, 6144
2. 1234, 4012, 5000, 6200
3. 1020, 3012, 6120, 8100
4. 2021, 4050, 5112, 7100

Solution: (C) 1020, 3012, 6120, 8100

Explanation:

Pages which are not present in the  main memory are:

 

Page	Address	Address that will cause page fault
1	1K	1024-2047
3	3K	3072-4095
4	4K	4096-5119
6	6K	6144-7167

 

1020 will not result in a page fault (1024-2047)

3012 does not result in a page fault (3072-4095)

Page faults will not be caused by 6120. (4096-5119)

8100 will not result in a page fault (6144-7167)

 

Question 2: Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one microsecond, but if a page fault occurs, it takes 2001 microseconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available?

1. 60 sec
2. 30 sec
3. 45 sec
4. 10 sec

Solution: (C) 45 sec

Explanation:

T = N_instr x 1µs + 15,000 x 2,000 µs = 60s

N_instr x 1µs = 60,000,000 µs – 30,000,000 µs = 30,000,000 µs

N_instr = 30,000,000

The number of instruction between two page faults is

N_instr /N_PageFaults = 30,000,000/15,000 = 2,000

When the mean interval between page faults is twice, the amount of instructions between two page faults is also doubled, resulting in a total of 4,000 instructions.

Now the number of page faults = 30,000,000/4,000 = 7,500

T’ = 30,000,000 µs + 7,500 x 2,000 µs

= 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s

Doubling the RAM does not imply that the software will run twice as fast as before. The performance boost is 25% in this case.

 

Question 3: Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. What is the time in msec to read a block of 1024 bytes from this disk?

1. 57 msec
2. 49 msec
3. 48 msec
4. 17 msec

Solution: (B) 49 msec

Explanation:

Seek time + average rotational delay + transfer time = time in milliseconds to read a block of 1024 bytes (Access time or Disk Latency).

There are 1024/16384 tracks to read for this block if each track is 16384 bytes long.

Seek time = 40 msec

Rotational delay = 16 msec

Transfer Time = (sectors_read/sectors per rev.) x rotational delay

               = (1024/16384) x 16 = 1

Average rotational delay = rotational delay/2 = 16/2 = 8

Access time = 40 + 8 + 1 = 49 

 

Question 4: A system has 4 processes and 5 allocatable resources. The current allocation and maximum needs are as follows:
 

	Allocated	Maximum	Available
Process A	1 0 2 1 1	1 1 2 1 3	0 0 x 1 1
Process B	2 0 1 1 0	2 2 2 1 0
Process C	1 1 0 1 0	2 1 3 1 0
Process D	1 1 1 1 0	1 1 2 2 1

The smallest value of x for which the above system is safe is .............

1. 1
2. 3
3. 2
4. Not safe for any value of x.

Solution: D

Explanation:

For     A =  0  1  0  0  2, B = 0  2  1  0  0, C = 1  0  3  0  0, D = 0  0  1  1  1 

available is  0  0  x  1  1  

So, only D can execute for any no of x  as A, B or C need at least one instance of either of first 2 resources

If x=1, then D can be executed and will release its allocated resource which is 1 1 1 1 0   then available matrix will become   1  1  2  2  1 but with this, neither of (A, B,C ) process can be executed as their need matrix is not less than or equal to the available matrix so it will result in an unsafe state.

If x=2, then D can be executed and will release its allocated resource which is 1 1 1 1 0   then the available matrix will become 1  1  3  2  1 now Process C (with need 1 0 3 0  0 ) can be executed and release its allocated resources (1 1 0 1  0 ) and then available matrix will become  2   2  3  3  1  so B can be executed and then A (assuming some minor mistake in the last resource)  hence it is a safe state.

So the answer should be  C,  x=2

But, process A can not be executed for any no of x  as its fifth resource need is 2 which can not be fulfilled. Hence no option is available for it, So, option (D) is correct.
 

Question 5: In Unix, the login prompt can be changed by changing the contents of the file ................

1. contrab
2. Init
3. gettydefs
4. inittab

Solution: (C) gettydefs

Explanation: 

• The contents of the file gettydefs can be altered to change the login prompt under Unix.
• contrab is unique in the Unix world.
• When a Unix system is rebooted or forced to change run levels, the inittab file determines what happens.
• The first process that runs when a computer system boots up is init (short for initialization).

As a result, option (C) is the correct answer.

 

Question 6: A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?

1. pn
2. p
3. (2n-1)(p+1)
4. (p+1)n

Solution: (D) = (p+1)n

Explanation:

pn the maximum number of distinct tuples that you can form with p distinct values per dimensions.

You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.

(2n-1)×(p+1) is the minimum number of cells possible in the data cube, C.

(p+1)n is the maximum number of cells possible (including both base cells and aggregate cells) in the data cube, C.

 

Question 7: Suppose that from given statistics, it is known that meningitis causes stiff neck 50% of the time, that the proportion of persons having meningitis is 1/50000, and that the proportion of people having stiff neck is 1/20. Then the percentage of people who had meningitis and complain about stiff neck is:

1. 0.01%
2. 0.02%
3. 0.04%
4. 0.05%

Solution: (B) 0.02%

Explanation:

The computation is based on the Bayes' formula, which has been simplified.

P{B|A} = (P{A|B}.P{B}) / P{A}

P{M|S} = chance that a person had meningitis, based on the presence of a stiff neck.

P{S|M} denotes the likelihood that a person will complain of stiff neck if they have meningitis = 1/2 = 50%

P{S} = proportion of people who complain about stiff neck. = 1/20

 P{M} = proportion of people who had meningitis = 1/150000.

After that, P{M|S} = (P{S|M}.P{M}) / P{S} = (1/2 x 1/50,000) / 1/20 = 0.0002 = 0.02%.

 

Question 8: ................. system is market oriented and is used for data analysis by knowledge workers including Managers, Executives and Analysts.

1. OLTP
2. OLAP
3. Data System
4. Market System

Solution: (B) OLAP

Explanation:

OLAP (online analytical processing) is a computer approach that allows users to extract and query data quickly and selectively in order to examine it from many perspectives. Trend analysis, financial reporting, sales forecasting, budgeting, and other planning tasks are frequently aided by OLAP business intelligence queries.
 

Question 9: .................. allows selection of the relevant information necessary for the data warehouse.

1. The Top-Down View
2. Data Warehouse View
3. Data source View
4. Business Query View

Solution: (A) The Top-Down View

Explanation:

There are four points of view on the design of a data warehouse.

• The top-down view enables for the selection of relevant data for the data warehouse.
• The data source view provides the data that operational systems acquire, store, and manage.
• Fact tables and dimension tables make up the data warehouse view.
• The business query view looks at data in the warehouse from the perspective of an end-user.
   

Question 10: The hash function used in double hashing is of the form:

1. h(k, i)=(h1(k)+h2(k)+i)mod m
2. h(k, i)=(h1(k)+h2(k)-i)mod m
3. h(k, i)=(h1(k)+ih2(k))mod m
4. h(k, i)=(h1(k)-ih2(k))mod m

Solution: (C) h(k, i)=(h1(k)+ih2(k))mod m

Explanation:

The hash function used in double hashing is of the form

h(i, k) = (h₁(k) + i*h₂(k)) mod |T|

Hence, C is the answer.
 

Question 11: In the following graph, discovery time stamps and finishing time stamps of Depth First Search (DFS) are shown as x/y where x is discovery time stamp and y is finishing time stamp.

It shows which of the following depth first forest?

1. {a,b,e} {c,d,f,g,h}
2. {a,b,e} {c,d,h} {f,g}
3. {a,b,e} {f,g} {c,d} {h} 
4. {a,b,c,d} {e,f,g} {h}

Solution: (A) {a,b,e} {c,d,f,g,h}

Explanation:

The cycles in the graph, you will find that node a, b, e from one cycle. Node f, g form another cycle. Node c, d form another cycle and h is having a self loop {a,b,e} {f,g} {c,d} {h}.

 

Question 12: The number of disk pages access in B-tree search, where h is height, n is the number of keys, and t is the minimum degree, is:

1. θ(log_n h*t)
2. θ(log_t n*h)
3. θ(log_h n)
4. θ(log_t n)

Solution: (D) θ(log_t n)

Explanation:

B-TREE-SEARCH accesses θ(h) =θ(log_t n) disc pages, where h is the height of the B-tree and n is the number of keys in the B-tree. As a result, Option 4 is right.
 

Question 13: The inorder traversal of the following tree is:

1. 2 3 4 6 7 13 15 17 18 18 20
2. 20 18 18 17 15 13 7 6 4 3 2
3. 15 13 20 4 7 17 18 2 3 6 18
4. 2 4 3 13 7 6 15 17 20 18 18

Solution: (D) 2 4 3 13 7 6 15 17 20 18 18

Explanation:

It signifies that after traversing the root node, the left subtree is visited first, followed by the right subtree. Inorder traversal refers to the process of traversing the root node between the left and right subtrees. As a result, each node in the inorder traversal gets visited in between its subtrees.
 

Question 14: An ideal sort is an in-place-sort whose additional space requirement is ...............

1. O(log₂n) 
2. O(nlog₂n)
3. O(1)
4. O(n)

Solution: (C) O(1)

Explanation:

In place sorts are those that do not require extra room and use their own space; if space is necessary, it is O. (1).

As a result, option (C) is the correct answer.
 

Question 15: Which of the following is not a congestion policy at the network layer?

1. Flow Control Policy
2. Packet Discard Policy
3. Packet Lifetime Management Policy
4. Routing Algorithm

Solution: (A) Flow Control Policy

Explanation:

The technique of controlling the rate of data transmission between two nodes in order to avoid a rapid transmitter from overwhelming a slow receiver is known as flow control. Congestion policies on the network layer include the Routing Algorithm, Packet Lifetime Management Policy, and Packet Discard Policy.

As a result, option (A) is the correct answer.
You can read related articles such as Congestion Control in Computer Networks here.

Question 16: Loop unrolling is a code optimization technique:

1. that avoids tests at every iteration of the loop
2. that improves performance by decreasing the number of instructions in a basic block.
3. that exchanges inner loops with outer loops
4. that reorders operations to allow multiple computations to happen in parallel.

Solution: (A) that avoids tests at every iteration of the loop

Explanation: 

Loop unrolling is a code optimization technique that eliminates the need for tests at each loop iteration. Reducing the amount of instructions in a basic block has no effect on performance. It doesn't swap inner and outer loops or rearrange operations to allow many computations to run in parallel.

As a result, option (A) is the correct answer.

 

Question 17: What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions?

Mov al, 15

Mov ah, 15

Xor al, al

Mov cl, 3

Shr ax, cl

1. 0F00 H
2. 0F0F H
3. 01E0 H
4. FFFF H

Solution: (C) 01E0 h

Explanation:

In assembly,  ax=ah+bl

al is the lower 8 bits

ah is the higher bits 8

Here al and ah both contains 15 or F so ax contains  0F0F  or 0000 1111 0000 1111  

XOR al will return 0 in also ax becomes 0000 1111 0000 0000

MOV CL,3 will store  3 in CL  

shr ax,cl  will shift right ax by 3       

so 0000 1111 0000 0000 will

become  0000 0001 1110 0000 or 01E0 in hexadecimal i.e C 01E0 H

 

Question 18: Which of the following statements is false?

1. Top-down parsers are LL parsers where the first L stands for left-to-right scan and the second L stands for a leftmost derivation.
2. (000)* is a regular expression that matches only strings containing an odd number of zeroes, including the empty string.
3. Bottom-up parsers are in the LR family, where L stands for left-to-right scan and R stands for rightmost derivation
4. The class of context-free languages is closed under reversal. That is, if L is any context-free language, then the language LR={WR:wϵL} is context-free.

Solution: (B) (000)* is a regular expression that matches only strings containing an odd number of zeroes, including the empty string.

Explanation:

• Top-down parsers are LL parsers, with the first L indicating a left-to-right scan and the second L indicating the leftmost derivation. The regular expression Correct (000)* matches only strings with an odd number of zeroes, including the empty string. This is erroneous because incorrect (000)* generates, 000, 000000, 000000000,........ multiples of 3 that contain odd and even strings.
• The LR family includes bottom-up parsers, where L stands for left-to-right scan and R represents for rightmost derivation.
• Correct
• Under reversal, the context – free languages class is closed. That is, if L is a context-free language, then LR = wR: wƐL is a context-free language. Correct

As a result, option (B) is the correct answer.

 

Question 19: System calls are usually invoked by using:

1. A privileged instruction
2. An indirect jump
3. A software interrupt
4. Polling

Solution: (C) A software interrupt

Explanation:

• A software interrupt is frequently used to call a system call.
• Polling is the process of a computer or controlling device waiting for an external device to check its readiness or condition, which is frequently done with low-level hardware.
• A privileged instruction (typically written in machine code) is one that can only be executed by the operating system in a specified mode.
• The target address (i.e. the relative offset value) is encoded into the jump instruction itself in direct jump.

As a result, option (C) is the correct answer.
 

Question 20: The ............... transfers the executable image of a C++ program from hard disk to main memory.

1. Compiler
2. Linker
3. Debugger
4. Loader

Solution: (D) Loader

Explanation:

• The loader copies a C++ program's executable image from the hard disc to main memory.
• A debugger is a piece of software that can be used to test and debug other programmes (target program).
• Compiler A compiler is a piece of software that turns a high-level language (Source Language) into a low-level language (Object/Target/Machine Language).
• Linker takes object files and links them together, as well as any library files you specify, to create an executable output file.

As a result, option (D) is the correct answer.
 

Question 21: In software testing, how the error, fault and failure are related to each other?

1. Error leads to failure but fault is not related to error and failure
2. Fault leads to failure but error is not related to fault and failure
3. Error leads to fault and fault leads to failure
4. Fault leads to error and error leads to failure

Solution: (C) Error leads to fault and fault leads to failure

Explanation:

Error leads to fault, and fault leads to failure in software testing. The remaining options are incorrect.

As a result, option (C) is the correct answer.

 

Question 22: Which of the following is not a software process model?

1. Prototyping
2. Iterative
3. Timeboxing
4. Glassboxing

Solution: (D) Glassboxing

Explanation:

Iterative prototyping and prototyping are well-known software process models. Development is done iteratively in the timeboxing model, just like in the iterative enhancement paradigm, but each iteration is done in a timebox with a fixed duration. A software testing pattern is the glass boxing model.

As a result, option (D) is correct.

 

Question 23: How many solutions are there for the equation x+y+z+u=29 subject to the constraints that x≥1, y≥2, z≥3 and u≥0?

1. 4960
2. 2600
3. 23751
4. 8855

Solution: (B) 

Explanation:

Lets take, y1=x-1, y2=y-2, y3=z-3, y4=u-0.

Count the no. of solutions of = y1+y2+y3+y4=29-6=23

n=4, r=23

No. of solutions is C(n+r-1, r) 

= C(4+23-1, 23)

            = C(26,23) = C(26,3) = 26x25x24/1x2x3 = 2600

 

Question 24: A unix file system has 1-KB blocks and 4-byte disk addresses. What is the maximum file size if i-nodes contain 10 direct entries and one single, double and triple indirect entry each?

1. 32 GB
2. 64 GB
3. 16 GB
4. 1 GB

Solution: (C) 16 GB

Explanation:

Block = 2¹⁰

Block pointer size = 4 byte

Entries possible in block = 2¹⁰/2² = 256

Direct pointer gives = 10 * 256 = 10 blocks

Single indirect gives = 256 * 256 = 2⁸ blocks

Double indirect gives = 256 * 256 * 256 = 2¹⁶ blocks

Triple indirect = 256 * 256 * 256 * 256 = 2²⁴ blocks

Total blocks = 10 blocks + 2⁸ blocks + 2¹⁶ blocks + 2²⁴ blocks

= 16843018 blocks = 16843018 * 1024 = 17247250432 ≈ 16 GB

 

Question 25: ................. uses electronic means to transfer funds directly from one account to another rather than by cheque or cash?

1. M-Banking
2. E-Banking
3. O-Banking
4. C-Banking

Solution: (B) E-Banking

Explanation:

Electronic banking, often known as Electronic Funds Transfer (EFT), is simply the use of electronic means to transfer payments from one account to another instead of using a check or cash.

 

FAQs

What is the UGC NET exam?

UGC NET is a national-level exam organized by UGC to determine the eligibility of the candidates for lectureship and JRF.

What is the exam format for the UGC NET?

The UGC NET exam is conducted entirely online.

Is it possible to take the UGC NET after completing a postgraduate degree?

You are qualified to take the exam if you have finished a PG Diploma from a recognized university.

How can solving PYQs help in my exam preparation?

Solving PYQs will give you a good idea about the exam pattern and help you identify your weak topics to prepare them better for the examination.

How many papers are there in the UGC NET exam?

There are two papers, and the candidates get 3 hours for both papers. There are 150 questions in UGC NET combining both papers.